Question

A particle travels with acceleration ~a = 2e−t ˆi + 5 cost jˆ − 3 sin tˆk.

If the initial (i.e. at time t=0) position of the particle is located at (1, −3, 2) and its initial velocity is 4ˆi − 3jˆ + 2ˆk, find its displacement at time t

Answer 1

Given acceleration, $\vec{a} = 2e^{-t}\hat{i}+5cost\hat{j}-3sint\hat{k}$

Velocity is given by, $\vec{v} = \int \vec{a} dt = \int (2e^{-t}\hat{i}+5cost\hat{j}-3sint\hat{k})dt$

Integrating, $\vec{v} = ( -2e^{-t} + a)\hat{i} + (5sint + b)\hat{j} - (-3cost + c)\hat{k}$

Since we are given velocity at t=0,

Then,

$4\hat{i}-3\hat{j}+2\hat{k} = ( -2e^{0} + a)\hat{i} + (5sin0 + b)\hat{j} - (-3cos0 + c)\hat{k}$

Comparing both sides, $-2+a = 4 \implies a = 6, b=-3, c=1$

Then, $\vec{v} = ( -2e^{-t} + 6)\hat{i} + (5sint -3)\hat{j} - (-3cost + 1)\hat{k}$

Displacement is given by, $\vec{s} = \int \vec{v} dt$

$\vec{s} = \int [( -2e^{-t} + 6)\hat{i} + (5sint -3)\hat{j} - (-3cost + 1)\hat{k}] dt$

Integrating, $\vec{s} = [( 2e^{-t} + 6t+d)\hat{i} + (-5cost -3t+f)\hat{j} - (-3sint + t+g)\hat{k}]$

d, f and g are constant.

We are given position at time t=0,

$1\hat{i}-3\hat{j}+2\hat{k} = [( 2e^{0} + 6\times0+d)\hat{i} + (-5cos0 -3\times0 +f)\hat{j} - (-3sin0 + 0+g)\hat{k}]$

Comparing both sides,

we get, $d=-1, f=2, g=-2$

Then Displacement will be,

$\vec{s} = ( 2e^{-t} + 6t-1)\hat{i} + (-5cost -3t+2)\hat{j} - (-3sint + t-2)\hat{k}$